\(\int F^{c (a+b x)} (f \text {sech}(d+e x))^n \, dx\) [95]

Optimal result

Integrand size = 20, antiderivative size = 88 \[ \int F^{c (a+b x)} (f \text {sech}(d+e x))^n \, dx=\frac {\left (1+e^{2 d+2 e x}\right )^n F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (n,\frac {1}{2} \left (n+\frac {b c \log (F)}{e}\right ),\frac {1}{2} \left (2+n+\frac {b c \log (F)}{e}\right ),-e^{2 d+2 e x}\right ) (f \text {sech}(d+e x))^n}{e n+b c \log (F)} \] Output:

(1+exp(2*e*x+2*d))^n*F^(c*(b*x+a))*hypergeom([n, 1/2*n+1/2*b*c*ln(F)/e],[1 
+1/2*n+1/2*b*c*ln(F)/e],-exp(2*e*x+2*d))*(f*sech(e*x+d))^n/(e*n+b*c*ln(F))
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.03 \[ \int F^{c (a+b x)} (f \text {sech}(d+e x))^n \, dx=\frac {\left (1+e^{2 (d+e x)}\right )^n F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (n,\frac {e n+b c \log (F)}{2 e},1+\frac {e n+b c \log (F)}{2 e},-e^{2 (d+e x)}\right ) (f \text {sech}(d+e x))^n}{e n+b c \log (F)} \] Input:

Integrate[F^(c*(a + b*x))*(f*Sech[d + e*x])^n,x]
 

Output:

((1 + E^(2*(d + e*x)))^n*F^(c*(a + b*x))*Hypergeometric2F1[n, (e*n + b*c*L 
og[F])/(2*e), 1 + (e*n + b*c*Log[F])/(2*e), -E^(2*(d + e*x))]*(f*Sech[d + 
e*x])^n)/(e*n + b*c*Log[F])
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.25, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {7271, 6017, 2689}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int F^{c (a+b x)} (f \text {sech}(d+e x))^n \, dx\)

\(\Big \downarrow \) 7271

\(\displaystyle \text {sech}^{-n}(d+e x) (f \text {sech}(d+e x))^n \int F^{c (a+b x)} \text {sech}^n(d+e x)dx\)

\(\Big \downarrow \) 6017

\(\displaystyle e^{-n (d+e x)} \left (e^{2 (d+e x)}+1\right )^n (f \text {sech}(d+e x))^n \int e^{d n+e x n} \left (1+e^{2 (d+e x)}\right )^{-n} F^{a c+b x c}dx\)

\(\Big \downarrow \) 2689

\(\displaystyle \frac {e^{-n (d+e x)+d n+e n x} \left (e^{2 (d+e x)}+1\right )^n F^{a c+b c x} (f \text {sech}(d+e x))^n \operatorname {Hypergeometric2F1}\left (n,\frac {e n+b c \log (F)}{2 e},\frac {e n+b c \log (F)}{2 e}+1,-e^{2 (d+e x)}\right )}{b c \log (F)+e n}\)

Input:

Int[F^(c*(a + b*x))*(f*Sech[d + e*x])^n,x]
 

Output:

(E^(d*n + e*n*x - n*(d + e*x))*(1 + E^(2*(d + e*x)))^n*F^(a*c + b*c*x)*Hyp 
ergeometric2F1[n, (e*n + b*c*Log[F])/(2*e), 1 + (e*n + b*c*Log[F])/(2*e), 
-E^(2*(d + e*x))]*(f*Sech[d + e*x])^n)/(e*n + b*c*Log[F])
 

Defintions of rubi rules used

rule 2689

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_. 
) + (g_.)*(x_)))*(H_)^((t_.)*((r_.) + (s_.)*(x_))), x_Symbol] :> Simp[G^(h* 
(f + g*x))*H^(t*(r + s*x))*((a + b*F^(e*(c + d*x)))^p/((g*h*Log[G] + s*t*Lo 
g[H])*((a + b*F^(e*(c + d*x)))/a)^p))*Hypergeometric2F1[-p, (g*h*Log[G] + s 
*t*Log[H])/(d*e*Log[F]), (g*h*Log[G] + s*t*Log[H])/(d*e*Log[F]) + 1, Simpli 
fy[(-b/a)*F^(e*(c + d*x))]], x] /; FreeQ[{F, G, H, a, b, c, d, e, f, g, h, 
r, s, t, p}, x] &&  !IntegerQ[p]
 

rule 6017

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sech[(d_.) + (e_.)*(x_)]^(n_.), x_Sym 
bol] :> Simp[(1 + E^(2*(d + e*x)))^n*(Sech[d + e*x]^n/E^(n*(d + e*x)))   In 
t[SimplifyIntegrand[F^(c*(a + b*x))*(E^(n*(d + e*x))/(1 + E^(2*(d + e*x)))^ 
n), x], x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&  !IntegerQ[n]
 

rule 7271

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*v^m)^ 
FracPart[p]/v^(m*FracPart[p]))   Int[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, 
x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&  !(Eq 
Q[v, x] && EqQ[m, 1])
 

Maple [F]

\[\int F^{c \left (b x +a \right )} \left (f \,\operatorname {sech}\left (e x +d \right )\right )^{n}d x\]

Input:

int(F^(c*(b*x+a))*(f*sech(e*x+d))^n,x)
 

Output:

int(F^(c*(b*x+a))*(f*sech(e*x+d))^n,x)
 

Fricas [F]

\[ \int F^{c (a+b x)} (f \text {sech}(d+e x))^n \, dx=\int { \left (f \operatorname {sech}\left (e x + d\right )\right )^{n} F^{{\left (b x + a\right )} c} \,d x } \] Input:

integrate(F^(c*(b*x+a))*(f*sech(e*x+d))^n,x, algorithm="fricas")
 

Output:

integral((f*sech(e*x + d))^n*F^(b*c*x + a*c), x)
 

Sympy [F]

\[ \int F^{c (a+b x)} (f \text {sech}(d+e x))^n \, dx=\int F^{c \left (a + b x\right )} \left (f \operatorname {sech}{\left (d + e x \right )}\right )^{n}\, dx \] Input:

integrate(F**(c*(b*x+a))*(f*sech(e*x+d))**n,x)
 

Output:

Integral(F**(c*(a + b*x))*(f*sech(d + e*x))**n, x)
 

Maxima [F]

\[ \int F^{c (a+b x)} (f \text {sech}(d+e x))^n \, dx=\int { \left (f \operatorname {sech}\left (e x + d\right )\right )^{n} F^{{\left (b x + a\right )} c} \,d x } \] Input:

integrate(F^(c*(b*x+a))*(f*sech(e*x+d))^n,x, algorithm="maxima")
 

Output:

integrate((f*sech(e*x + d))^n*F^((b*x + a)*c), x)
 

Giac [F]

\[ \int F^{c (a+b x)} (f \text {sech}(d+e x))^n \, dx=\int { \left (f \operatorname {sech}\left (e x + d\right )\right )^{n} F^{{\left (b x + a\right )} c} \,d x } \] Input:

integrate(F^(c*(b*x+a))*(f*sech(e*x+d))^n,x, algorithm="giac")
 

Output:

integrate((f*sech(e*x + d))^n*F^((b*x + a)*c), x)
 

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} (f \text {sech}(d+e x))^n \, dx=\int F^{c\,\left (a+b\,x\right )}\,{\left (\frac {f}{\mathrm {cosh}\left (d+e\,x\right )}\right )}^n \,d x \] Input:

int(F^(c*(a + b*x))*(f/cosh(d + e*x))^n,x)
 

Output:

int(F^(c*(a + b*x))*(f/cosh(d + e*x))^n, x)
 

Reduce [F]

\[ \int F^{c (a+b x)} (f \text {sech}(d+e x))^n \, dx=f^{a c +n} \left (\int f^{b c x} \mathrm {sech}\left (e x +d \right )^{n}d x \right ) \] Input:

int(F^(c*(b*x+a))*(f*sech(e*x+d))^n,x)
 

Output:

f**(a*c + n)*int(f**(b*c*x)*sech(d + e*x)**n,x)